# Prove: cos4x = 8cos^4x - 8cos^2x + 1 My Attempt: RS: = 4cos^2x (2cos^2x - 1) + 1 = 4 cos^2x (cos2x) + 1 LS: = cos2(2x) = 2cos^2(2x) - 1 = (cos^2(2)) - cos^2(2x)) - 1 ----- Prove: 8cos^4x = cos4x + 4cos2x + 3 My Attempt: RS: = cos2(2x) + 4cos(

Prove:
cos4x = 8cos^4x - 8cos^2x + 1

My Attempt:
RS:
= 4cos^2x (2cos^2x - 1) + 1
= 4 cos^2x (cos2x) + 1

LS:
= cos2(2x)
= 2cos^2(2x) - 1
= (cos^2(2)) - cos^2(2x)) - 1

-----

Prove:
8cos^4x = cos4x + 4cos2x + 3

My Attempt:
RS:
= cos2(2x) + 4cos(2x) + 3
= 2cos^2(2x) - 1 + 2(4)cos^2(2x) - 1 + 3

Do you mean... why isn't (cos(2x))^2 equal to cos(4x^2)?
= (4cos^2(x)-2)(2cos^2(x)-1)
= (8cos^4(x) + 2) (-4cos^2(x)+2)
verify identity
cos 2A sec A = 2cos A - sec A
Since you square the entire function, why is it (cos2x)(cos2x)? I mean, why don't you square the "2" into (cosx)(cos2x)?
All these numbers and things are getting confusing. Let's try to simplify this.

Could you tell me how you foiled this?
(4cos^2(x)-2)(2cos^2(x)-1)
No, it would be much harder. It's always easier to expand than to condense.

Expansion of the identities is a very important concept.

cos(2x) = 2cos^2(x) - 1
If it's cos(4x), then that's like cos(2*2x). Everything in the expansion is the same except the x will be 2x now, like 2cos^2(2x) - 1.

All you're doing is repeating the expansion process until you can't do it anymore.

If you have more questions that are confusing you, then I can try to help.
Whenever I try to prove the identity with the right side, I keep getting different answer

RS:
= cos2(2x) + 4(2cos^2(x)-1) + 3
= cos2(2x) + 8cos^2(x) - 4 + 3
= 2cos^2(2x) - 1 + 8cos^2(x) - 4 + 3
= 2cos^2(2x) + 8cos^2(x) - 2

...I think I'm making this too complicated than it should be...
Made a mistake on the last line

= (4cos(2x))*(2cos^2(2x) - 1)*(2cos^2(2x) - 1) * (cos(2x))
Ah okay, I get it now, so "2x" is like "x", they are always together

I'm working on the second question now and I'm expanding the identity, but I'm kind of stuck...

8cos^4x = cos4x + 4cos2x + 3

LS:
= 8cos^4x
= 8(cos^2x)(cos^2x)
= 4(cos^4(2x))
= 4(cos^2(2x))*(cos^2(2x))
= 4(cos2x)*(cos2x)*(cos2x)*(cos2x)
= (4cos^2(2x))*(2cos^2(2x) - 1)*(2cos^2(2x) - 1) * (cos(2x))
Ugh...nevermind I'm so confused...

If I did the right side, would it be more easy than the left side? I don't think I know how to expand with identities...
I'm working on it...
I'll try to explain the two lines better...

2cos^2(2x) - 1
This is like 2(cos^2(2x)) - 1.

In this line, we're changing only the cos^2(2x). When the squared is written on the "cos" part, that means the entire function is squared. Therefore, cos^2(2x) is the same as (cos(2x))^2.

When something is squared, you multiply it by itself. For example, 4^2 = 4*4, and (n+1)^2 = (n+1)(n+1).

Therefore, we are saying that (cos(2x))^2 = (cos(2x))(cos(2x)) = cos2x * cos2x. The rest of the line remained unchanged.
And where did you get the "- 2" from 2cos^2(2x) + 8cos^2(x) - 2
And I still get stuck...

= cos2(2x) + 4(2cos^2(2x)-1) + 3
= cos2(2x) + 8cos^2(2x) - 4 + 3
= 2cos^2(2x) - 1 + 8cos^2(2x) - 4 + 3
= 2cos^2(2x) + 8cos^2(2x) - 2

What am I doing wrong now?
Okay? But, I still don't get what you did in:

2cos^2(2x) - 1 Double-angle again.
2(cos2x * cos2x) - 1 Double-angle again.
##### Add Answer of: Prove: cos4x = 8cos^4x - 8cos^2x + 1 My Attempt: RS: = 4cos^2x (2cos^2x - 1) + 1 = 4 cos^2x (cos2x) + 1 LS: = cos2(2x) = 2cos^2(2x) - 1 = (cos^2(2)) - cos^2(2x)) - 1 ----- Prove: 8cos^4x = cos4x + 4cos2x + 3 My Attempt: RS: = cos2(2x) + 4cos(
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